x is equal to the product of the first n positive integers

IMO B

Given x = n!. The question is asking the minimum value of n such that x has 6 terminating zeros. The prime factors of such a number will surely include 2 & 5 (product of 10). Since every alternate number is even, lets check for the # of 5s in each of the cases that will tell us the number of trailing zeros:

A. 20/5 = 4, 4 trailing zeros
B. 25/5 = 5, 5/5=1 ==> (5+1) 6 training zeros
C. 30/5 = 6, 6/5=1 ==> (6+1) 7 trailing zeros
D. 35/5 = 7, 7/5=1 ==> (7+1) 8 trailing zeros
E. 40/5 = 8, 8/5=1 ==> (8+1) 9 trailing zeros

figs wrote:if the number x is equal to the product of the first n positive integers (1,2,3,...,n), what is the minimum value of n such that the last six digits of x are equal to 0?

A. 20
B. 25
C. 30
D. 35
E. 40


OA:after some explanation

If a number must have the last 6 digits as 0, then that number must have 10^6=2^6 x 5^6 as factor.
So let n!= G x 10^6. G contains factors of 2 and 5 of the form 2^m-6 and 5^n-6, where m is number of 2 and n number of 5 in the expression. So G will contain the leftover 2's and 5's, since we are only looking for the minimum.

20! has 10 even numbers so clearly m>= 10 but n is < 6. So NO.
30! has 15 even numbers and m=6, ( 30, 25, 20, 15, 10, 5), So it is the smallest that meets the parameters above.

Choose B.

dtweah wrote:

figs wrote:if the number x is equal to the product of the first n positive integers (1,2,3,...,n), what is the minimum value of n such that the last six digits of x are equal to 0?

A. 20
B. 25
C. 30
D. 35
E. 40


OA:after some explanation

If a number must have the last 6 digits as 0, then that number must have 10^6=2^6 x 5^6 as factor.
So let n!= G x 10^6. G contains factors of 2 and 5 of the form 2^m-6 and 5^n-6, where m is number of 2 and n number of 5 in the expression. So G will contain the leftover 2's and 5's, since we are only looking for the minimum.

20! has 10 even numbers so clearly m>= 10 but n is < 6. So NO.
30! has 15 even numbers and m=6, ( 30, 25, 20, 15, 10, 5), So it is the smallest that meets the parameters above.

Choose B.

I meant Choose C 30! Didn't consider 25! because it has only one more 5 than 20! and is still less than 6

Ian Stewart wrote:

dtweah wrote: I meant Choose C 30! Didn't consider 25! because it has only one more 5 than 20! and is still less than 6

25! has two more 5's than 20!, not one, since 25 = 5^2.

Vemuri has the right answer above; in the prime factorization of 25!, we have six 5's -- two from the 25, and one each from 20, 15, 10 and 5 -- and lots of 2's, so 25! is divisible by 10^6, and therefore ends in six zeros.

Yes, Ian one of those costly errors!!

thanks for the explanation

oa:b

I didn't understand the solution for this problem..

I just stared at the question and solution for 15 mins and still fail to understand it..

Can you guys please expand??

Thanks much in advance